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\(\frac{2000\cdot2001-1001}{1999\cdot2002-999}=\frac{1999\cdot2001+2001-1001}{1999\cdot2001+1999-999}\)
\(=\frac{1999\cdot2001+1000}{1999\cdot2001+1000}=1\)
Ta có :
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(A=3.\frac{6}{25}\)
\(A=\frac{18}{25}\)
Vậy \(A=\frac{18}{25}\)
Chúc bạn học tốt ~
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)
\(=\frac{3.4.6}{25.4}\)
\(\Rightarrow A=\frac{18}{25}\)
\(x\cdot\left(1990+2000+2001\right)=99\cdot7-99\cdot4-99-99\cdot2\)
\(\Leftrightarrow x\cdot\left(1999+2000+2001\right)=99\cdot\left(7-4-1-2\right)\)
\(\Leftrightarrow x\cdot\left(1999+2000+2001\right)=99\cdot0\)
\(\Leftrightarrow x\cdot\left(1999+2000+2001\right)=0\)
\(\Rightarrow x=0\)
\(x\times\left(1999+2000+2001\right)=99\times7-99\times4-99-99\times2\)
\(x\times\left(1999+2000+2001\right)=99\times\left(7-4-1-2\right)\)
\(x\times\left(1999+2000+2001\right)=99\times0\)
\(x\times\left(1999+2000+2001\right)=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
57x36+114x32-1999-2001
=57x36+57x2x32-(1999+2001)
=57x 36+57x64-4000
=57x(36+64)-4000
=57x100-4000
=5700-4000
=5300
57 × 36 + 114 × 32 - 1999 - 2001
= 57 × 36 + 57 × 2 × 32 - (1999 + 2001)
= 57 × 36 + 57 × 64 - 4000
= 57 × (36 + 64) - 4000
= 57 × 100 - 4000
= 5700 - 4000
= 1700
57 x 36 + 114 x 32 - 1999 - 2001
= 57 x 36 + 57 x 2 x 32 - ( 1999 + 2001 )
= 57 x 36 + 57 x 64 - 4000
= 57 x ( 36 + 64 ) - 4000
= 57 x 100 - 4000
= 5700 - 4000
= 1700